# Anyone good at physics?



## Sandstone-Shadow (Mar 25, 2010)

I'm currently taking AP Physics C (if that means anything to anyone; if it doesn't, it's an advanced physics class basically) and struggling with it. It's an early class which starts before everyone else's school day, so it's difficult to concentrate to begin with, and my teacher's teaching style does not help me at all. I'm going to explain how the class is set up because I think that's part of why I can't keep up with it; it's just such a detrimental structure for me. We have online homework assignments pretty much every night that we print off (which, by the way, leaves _very_ little room for writing, a fact that is infuriating because I need space to organize my work) and work on. We submit the answers online as well; it tells you instantly if the numbers are right or wrong, but it does not give you any sort of hint if you got the wrong answer; it just spits a red x in your face. 

We go over some questions in class, but pretty much only the day before it's due (all assignments are due at midnight the day we would go over problems) but this requires having done the assignments ahead of time so you know what problems are hard and require going-over. Working ahead is something that just doesn't work for me in this class; there's too much to do. There is a forum component to the online homework, but this is ridiculous since the teacher preaches to us to use the forum, but then she rarely checks it herself and when she does, her "helpful replies" are confusing and very, very concise.

Now, she has started pulling us out of our study halls into physics class to do assignments if we didn't get a 100% on them that night. I use my study hall for other homework; I'm not getting the other homework done at night because I'm working on physics that I don't understand, and I can't work on it during study hall because I'm being called out for physics that I _still_ don't understand.

Moral of this huge post: is anyone good at physics that would be willing to help me out? We're doing angular motion, energy, and momentum right now. I really need someone who can help explain some of this, or at least someone who can give me some tips so I can keep up with this class better.


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## shadow_lugia (Mar 25, 2010)

I _might_ be able to help with a couple of things, but probably not. I'm only getting into high school next year, and have never studied physics except in my free time from astrophysics books, which I enjoy reading. So, I doubt I could answer any questions unless they involve, say, black holes or something, but I may as well try to help.


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## Sandstone-Shadow (Mar 25, 2010)

We're doing stuff like calculating moments of inertia and torque and angular velocities and stuff like that. So if you know anything about that... if not, thanks for the attempt, I appreciate it. =)


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## shadow_lugia (Mar 25, 2010)

Oh... No, I'm not any good at that, sorry. Usually when I see "physics" I think "atoms," which I kind of get. Depending on what torque is, I might know something (probably not, since I have no idea what torque is :P). Otherwise, sorry for not being able to help.


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## Sandstone-Shadow (Mar 25, 2010)

Thanks anyway! =)


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## Minnow (Mar 25, 2010)

If you'd posted this earlier, then I'd probably suggest dropping it, if the work really was too much and you wouldn't lose much by dropping it (I don't know if you're taking it, basically for fun, or because of some physics requirement thing you're going for), but since the year is almost over now, I think it's best to try to tough it out. Perhaps you could talk to the teacher a bit about the problems you're having?

How do the other students seem to handle it?


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## Sandstone-Shadow (Mar 26, 2010)

The teacher doesn't really ever help. =/ She explains things but the way she explains is so confusing and crooked. She makes mistakes all the time, too; it's hard to really listen to her explanations because you also have to be checking to make sure she's not making a mistake. Everyone in the class pretty much agrees with me. =/

Ha yeah I wish I had dropped it though.


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## Murkrow (Mar 26, 2010)

I might be able to help you. With bits of it at least, I might not have done the same things that you are. (I wasn't there when we did torque in physics but I got the question right in the exam so I think I'm okay at that)

What sort of questions are they, like what information is usually given and are they usually in diagrams or sentences?


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## Kali the Flygon (Mar 26, 2010)

Is anyone good at physics? Boy are you lucky...

I'm currently working on my doctorate in physics, and could be done this year. I already have a bachelors and masters. I suppose that would qualify me as 'good'.

How much math have you had also, by the way? And do you have a way I can talk more real-time with you, like MSN messenger or something?


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## Lorem Ipsum (Mar 26, 2010)

I was going to offer my help, but then a Ph.D student turned up and stole my thunder! No worries, I'm only up to GCSE level Physics anyway, which probably wouldn't help you.


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## Sandstone-Shadow (Mar 28, 2010)

Awesome! I'm currently in Calculus AB, but I feel that as far as math goes I'm fine, unless calculus deals with some physics stuff as you progress in it. The problems are usually word problems, such as this one:



> A particular tower clock similar to Big Ben has an hour hand 2.70 m long with a mass of 51.0 kg, and a minute hand 4.20 m long with a mass of 100 kg (Figure P10.40). Calculate the total rotational kinetic energy of the two hands about the axis of rotation. (You may model the hands as long thin rods.)


So what I did is I used KE=.5Iω² for each clock hand and added them together, using I=(1/3)ml² and ω for the minute hand = 2π radians/60 minutes, and for the hour hand, ω = 2π radians/720 minutes. This didn't get me the right answer, though, so I guess I must be missing something.


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## Flitterbie (Mar 29, 2010)

I=(1/3)ml² only works when the rotation is around the center of mass. Since the axis is on the end, you want to use I=(1/12)ml².

At least, that's what I remember. Also, I'm pretty sure time for the hour hand would be 720 minutes, since it takes 12 hours to complete a rotation.


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## Sandstone-Shadow (Mar 29, 2010)

Hmm, our book taught us the other way around, although I'm not entirely trusting of my book. I tried it that way though; it didn't work. =(

Ah, right, I did use 720 minutes; I reduced it later and put the wrong number up there.


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## Kali the Flygon (Mar 29, 2010)

To be honest, what you wrote there now looks right. Are you sure you didn't make an error in the mathematics of your calculation? 

Yes, it's 1/3 for a rod about the end, and 1/12 for a rod about the middle. One way you can check that is by imagining a rod about the middle as two independent rods about the end.


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## Sandstone-Shadow (Mar 29, 2010)

Pretty sure. =/ I tried it a few times and got the same (wrong, apparently) answer each time. Is it fine to use radians/minute for the angular velocity? I'm never sure exactly when to use which units.

Hmm, okay. So if a rod about the middle was two independent rods rotating about the end, their lengths would be half... so (1/3)m(.5l)² = (1/12)ml². But wouldn't you have two of those rods? How come it isn't (1/6)ml²?


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## Flitterbie (Mar 30, 2010)

Kali the Flygon said:


> To be honest, what you wrote there now looks right. Are you sure you didn't make an error in the mathematics of your calculation?
> 
> Yes, it's 1/3 for a rod about the end, and 1/12 for a rod about the middle. One way you can check that is by imagining a rod about the middle as two independent rods about the end.


...Dammit, why do I get those two mixed up so often?

Typically, angular speed is measured in radians/second, though, as long as you're consistent, it shouldn't matter. But you'd likely want to use whatever units the book gives the answer in.


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## Sandstone-Shadow (Mar 30, 2010)

The answer is supposed to be in joules at the end. And joules are (kg m²)/s²... ohh, my velocity should be in m/s then, shouldn't it?

Edit: Still didn't work.


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## Flitterbie (Mar 30, 2010)

What's the book's answer?


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## Sandstone-Shadow (Mar 30, 2010)

I don't know. The assignments are online; they use problems from the book but change all the numbers so you can't just submit the book's answer. =/


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## Flitterbie (Mar 30, 2010)

Then how do you know you have the wrong answer?


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## Kali the Flygon (Mar 30, 2010)

Sandstone-Shadow said:


> Hmm, okay. So if a rod about the middle was two independent rods rotating about the end, their lengths would be half... so (1/3)m(.5l)² = (1/12)ml². But wouldn't you have two of those rods? How come it isn't (1/6)ml²?


While you have two, which doubles the mass, you also have to add the two separate inertias together, which doubles the 1/6 to a 1/3.

I was thinking it was possibly a matter of units, but if you're saying it still doesn't work, I'm getting confused. Perhaps you can send me your step by step work in a PM, and I can check it myself.


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## opaltiger (Mar 30, 2010)

I can contribute nothing to the physics, but if you've checked your work multiple times and keep getting the same result, the problem probably isn't in your end. Textbooks get solutions wrong all the time, and I have even less faith in online quizzes.


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## Sandstone-Shadow (Apr 1, 2010)

So it turns out the teacher thought I was doing it right too and just gave me credit for the problem... hrr.



> > Hmm, okay. So if a rod about the middle was two independent rods rotating about the end, their lengths would be half... so (1/3)m(.5l)² = (1/12)ml². But wouldn't you have two of those rods? How come it isn't (1/6)ml²?
> 
> 
> While you have two, which doubles the mass, you also have to add the two separate inertias together, which doubles the 1/6 to a 1/3.


But isn't the inertia for each of them (1/12)ml²? Or wait, it would actually be (1/12)(1/2)ml² since the mass would be half too, which would equal (1/24)ml², so then if you have two of those, you'd get (1/12)ml². Ahh, I understand that now. Thanks! =D

Alright, we started something new and the teacher hardly explained it at all, and basically shot me down when I tried to ask about it. How would I do a problem like this?



> Given M = 2 i + 2  j - 6 k and N = i  - 6j - k, calculate the vector product M multiplied by N.


We learned how to calculate cross products, but only for vectors with just i and j components, not i, j, and k. And the answer is looking for a vector with i, j, and k components. I'm confused, not only because I don't know how to take the cross product of something with three terms, but I'm also confused because when taking the cross product of vectors with just i and j components, you end up with just k components. Now apparently the i and j stay in there too?


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## opaltiger (Apr 1, 2010)

First make a 3x3 matrix where the top row consists of the unit vectors and the other two rows of the coefficients in both vectors (so second row is 2, 2, -6 and third is 1, -6, -1).

Then calculate the determinant of that matrix. It's as easy as that. If you don't know how to do that, I'll try and explain, though it's a little difficult in words.

Cover the first column (the i column). Then find the determinant of the 2x2 matrix which is left (if you ignore the j and k elements). Multiply this by i. Next cover the j column and find the determinant of the new 2x2 matrix formed from the numerical elements. Multiply this by j and subtract it from the first result. Finally, cover the third column, calculate the 2x2 determinant, multiply by k, and add to the first two.

Full disclosure: instead of subtracting the j result, you can reverse the order of the two columns and add it instead. But I find this way easier.


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## Kali the Flygon (Apr 1, 2010)

To say the same thing opaltiger said in a similar, but slightly different way, if you know how to do cross products of 2-component vectors, you can do them for 3-component vectors in the following way:

The cross product of two 3-component vectors is another 3-component vector, _X_.

Remove the _i_ components of the two vectors, and do the cross product between just the _j_ and _k_ components. The answer will be the _i_ component of _X_. Then, remove the _j_ components of the two initial vectors, and, importantly... switch the order of the _k_ and _i_ components, so that _k_ comes before _i_. Then do the cross product with just the _k_ and _i_ components, and write the answer as the _j_ component of _X_. Finally, remove the _k_ components of the two initial vectors and perform the cross product between the _i_ and _j_ components, and write the answer as the _k_ component of _X_.

In purely mathematical speak, if:
_M_ = a_i_ + b_j_ + c_k_,
_N_ = d_i_ + e_j_ + f_k_, then

_X_ = _M_x_N_ = (b*f - c*e)_i_ + (c*d - a*f)_j_ + (b*d - a*e)_k_.

Hope this helps.


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## Sandstone-Shadow (Apr 8, 2010)

Ah, okay... that makes sense, thanks guys! =)

Ugh, I'm stuck again. We're studying angular momentum, and I've tried reading over my notes from class, looking at the book, and looking at various other webpages. What I'm figuring out is that angular momentum (for, I'm assuming, a point) is r (the position vector) crossed with p (the linear momentum) so you end up with L = r X mv.

And I also found that for a system of particles, angular momentum is L = Iω.

First of all, is what I posted correct?

And second of all, how do I use those things to solve something like: 



> The position vector of a particle of mass 1.70 kg is given as a function of time by r = (6.00 i + 5.10 t j) m. Determine the angular momentum of the particle about the origin, as a function of time.


For this one, L = (6i + 5.1tj) X 1.7v. (Right?) But it doesn't give you v at all; this seems like a really basic concept to me but I can't figure it out.

I'm running into similar issues in other problems, but the other problems are all more complicated than the one that I posted, argh.


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## Kali the Flygon (Apr 10, 2010)

Yes, those equations look correct, though it's I x ω, as opposed to other forms of products. I assume that's what you meant.

The velocity is related to the change in the radius with respect to time, in particular v = dr/dt.


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## Sandstone-Shadow (Apr 13, 2010)

Kali the Flygon said:


> Yes, those equations look correct, though it's I x ω, as opposed to other forms of products. I assume that's what you meant.


Hmm, okay. Our teacher told us it was just Iω, because for some "pure rotation" thing the angle between the position vector r and the velocity v is always 90 degrees? I really wasn't sure what she meant though, and she never explained it or explained what "pure rotation" is. Is there any truth to what she's saying?



> The velocity is related to the change in the radius with respect to time, in particular v = dr/dt.


Ah... awesome, I finally got one right! =D Thank you so much for your help so far, I really appreciate it.


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## Kali the Flygon (Apr 14, 2010)

Well, I suppose you're right about Iω, considering the direction of the inertia is just the axis of rotation, and the angular velocity measures in the perpendicular direction to that axis always.

Glad I could help... even if it's just to get you to write things out and spur your own thoughts... *references visitor message*


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